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3 years ago

6

A person sitting in a chair with wheels stands up, causing the chair to roll backward across the floor. How would you describe t

he momentum of the chair?

1 answer:

OleMash [197]3 years ago

3 0

Scenes the chair wheels are up the person is rolling backwards and if the wheels were down then the person would go forwards
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Las 3 leyes de klepler del sistema planetario<br><br><br><br> AYUDAA¡¡

shtirl [24]

I actually don't know. i speak english.

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3 years ago

Which of the following is not true?

ahrayia [7]

The answer is either C or D..

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3 years ago

A load of 800 N is lifted using a block and tackle having 5 pulleys. If the applied effort is 200 N, calculate

ser-zykov [4K]

Explanation:

Load=800N

Effort=200N

1. Mechanical Advantage = LOAD/EFFORT

= 800N/200N

= 4

2 Velocity Ratio = no. Of pulleys =5

3. Efficiency = Mechanical advantage / velocity ratio × 100%

= (4/5)×100%

=80%

4. output work= load×load distance

= 800N × 5m

= 4 × 1000J

5. Efficiency = (output work/input work) ×100%

Or, 80% = (4000J/input work) ×100%

Or, 80%/100% = 4000J/inputwork

Or, 4/5 = 4000J/inputwork

Or, input work =4000J × 5/4

Input work = 5×1000J

I hope it helped! ;-)

8 0

3 years ago

Read 2 more answers

Suppose that a comet that was seen in 550 A.D. by Chinese astronomers was spotted again in year 1941. Assume the time between ob

pickupchik [31]

To solve the problem it is necessary to apply the concepts related to Kepler's third law as well as the calculation of distances in orbits with eccentricities.

Kepler's third law tells us that

$T^2 = \frac{4\pi^2}{GM}a^3$

Where

T= Period

G= Gravitational constant

M = Mass of the sun

a= The semimajor axis of the comet's orbit

The period in years would be given by

$T= 1941-550\\T= 1391y(\frac{31536000s}{1y})\\T=4.3866*10^{10}s$

PART A) Replacing the values to find a, we have

$a^3= \frac{T^2 GM}{4\pi^2}$

$a^3 = \frac{(4.3866*10^{10})^2(6.67*10^{-11})(1.989*10^{30})}{4\pi^2}$

$a^3 = 6.46632*10^{39}$

$a = 1.86303*10^{13}m$

Therefore the semimajor axis is $1.86303*10^{13}m$

PART B) If the semi-major axis a and the eccentricity e of an orbit are known, then the periapsis and apoapsis distances can be calculated by

$R = a(1-e)$

$R = 1.86303*10^{13}(1-0.997)$

$R= 5.58*10^{10}m$

7 0

3 years ago

The electric field between the plates of the cathode ray tube of an older television set can be as high as 2.5x104 N/C. Determin

pentagon [3]

Answer:

(a) $F = -4.01 * 10^{-15} N$

(b) $a = 4.40 * 10^{15} m/s^2$

Explanation:

Parameter given:

Electric field, E = $2.5 * 10^4 N/C$

(a) Electric force is given (in terms of electric field) as a product of electric charge and electric field.

Mathematically:

$F = qE$

Electric charge, q, of an electron = $- 1.602 * 10^{-19} C$

$F = -1.602 * 10^{-19} * 2.5 * 10^4\\\\\\F = -4.01 * 10^{-15} N$

(b) This electrostatic force causes the electron to accelerate with an equivalent force:

F = -ma

where m = mass of an electron

a = acceleration of electron

(Note: the force is negative cos the direction of the force is opposite the direction of the electron)

Therefore:

$-ma = -4.01 * 10^{-15} N\\\\\\a = \frac{-4.01 * 10^{-15}}{-m}$

Mass, m, of an electron = $9.11 * 10^{-31} kg$

=> $a = \frac{-4.01 * 10^{-15}}{-9.11 * 10^{-31}}\\\\\\a = 4.40 * 10^{15} m/s^2$

The acceleration of the electron is $4.40 * 10^{15} m/s^2$

5 0

3 years ago